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Theorem List for Metamath Proof Explorer - 27801-27900   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremconimpfalt 27801 Assuming a, not b, and a implies b, there exists a proof that a is false.) (Contributed by Jarvin Udandy, 29-Aug-2016.)
 |-  ph   &    |-  -.  ps   &    |-  ( ph  ->  ps )   =>    |-  ( ph  <->  F.  )
 
Theoremaistbisfiaxb 27802 Given a is equivalent to T., Given b is equivalent to F. there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  ( ph 
 <->  T.  )   &    |-  ( ps  <->  F.  )   =>    |-  ( ph  \/_  ps )
 
Theoremaisfbistiaxb 27803 Given a is equivalent to F., Given b is equivalent to T., there exists a proof for a-xor-b. (Contributed by Jarvin Udandy, 31-Aug-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   =>    |-  ( ph  \/_  ps )
 
Theoremabcdta 27804 Given (((a and b) and c) and d), there exists a proof for a (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |-  ph
 
Theoremabcdtb 27805 Given (((a and b) and c) and d), there exists a proof for b (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 ps
 
Theoremabcdtc 27806 Given (((a and b) and c) and d), there exists a proof for c (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 ch
 
Theoremabcdtd 27807 Given (((a and b) and c) and d), there exists a proof for d (Contributed by Jarvin Udandy, 3-Sep-2016.)
 |-  (
 ( ( ph  /\  ps )  /\  ch )  /\  th )   =>    |- 
 th
 
Theoremmdandyv0 27808 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  F.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv1 27809 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  F.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv2 27810 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  F.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv3 27811 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  F.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv4 27812 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  T.  )   &    |-  ( et 
 <->  F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv5 27813 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  T.  )   &    |-  ( et 
 <->  F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv6 27814 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  T.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv7 27815 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  T.  )   &    |-  ( et  <->  F.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ph ) )
 
Theoremmdandyv8 27816 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  F.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv9 27817 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  F.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv10 27818 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  F.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv11 27819 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  F.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ph ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv12 27820 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  T.  )   &    |-  ( et 
 <->  T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv13 27821 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  T.  )   &    |-  ( et 
 <->  T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <-> 
 ph ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv14 27822 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  F.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  T.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ph )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyv15 27823 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ph, ps accordingly (Contributed by Jarvin Udandy, 6-Sep-2016.)
 |-  ( ph 
 <->  F.  )   &    |-  ( ps  <->  T.  )   &    |-  ( ch 
 <->  T.  )   &    |-  ( th  <->  T.  )   &    |-  ( ta 
 <->  T.  )   &    |-  ( et  <->  T.  )   =>    |-  ( ( ( ( ch  <->  ps )  /\  ( th 
 <->  ps ) )  /\  ( ta  <->  ps ) )  /\  ( et  <->  ps ) )
 
Theoremmdandyvr0 27824 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr1 27825 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr2 27826 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr3 27827 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr4 27828 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr5 27829 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr6 27830 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr7 27831 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  ze ) )
 
Theoremmdandyvr8 27832 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr9 27833 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr10 27834 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr11 27835 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  ze ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr12 27836 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr13 27837 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <->  ze ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr14 27838 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  ze )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvr15 27839 Given the equivalences set in the hypotheses, there exist a proof where ch, th, ta, et match ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph 
 <->  ze )   &    |-  ( ps  <->  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  <->  si )  /\  ( th 
 <-> 
 si ) )  /\  ( ta  <->  si ) )  /\  ( et  <->  si ) )
 
Theoremmdandyvrx0 27840 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx1 27841 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx2 27842 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  ze )
 )
 
Theoremmdandyvrx3 27843 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  ze ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx4 27844 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx5 27845 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx6 27846 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx7 27847 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ph )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  si ) )  /\  ( et  \/_  ze ) )
 
Theoremmdandyvrx8 27848 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx9 27849 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx10 27850 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_ 
 ze ) )  /\  ( et  \/_  si )
 )
 
Theoremmdandyvrx11 27851 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ph )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  ze ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx12 27852 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx13 27853 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ph )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  ze )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx14 27854 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ph )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  ze )  /\  ( th  \/_  si )
 )  /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
Theoremmdandyvrx15 27855 Given the exclusivities set in the hypotheses, there exist a proof where ch, th, ta, et exclude ze, si accordingly (Contributed by Jarvin Udandy, 7-Sep-2016.)
 |-  ( ph  \/_  ze )   &    |-  ( ps  \/_  si )   &    |-  ( ch  <->  ps )   &    |-  ( th  <->  ps )   &    |-  ( ta  <->  ps )   &    |-  ( et  <->  ps )   =>    |-  ( ( ( ( ch  \/_  si )  /\  ( th  \/_  si ) ) 
 /\  ( ta  \/_  si ) )  /\  ( et  \/_  si ) )
 
TheoremH15NH16TH15IH16 27856 Given 15 hypotheses and a 16th hypothesis, there exists a proof the 15 imply the 16th. (Contributed by Jarvin Udandy, 8-Sep-2016.)
 |-  ph   &    |-  ps   &    |-  ch   &    |-  th   &    |-  ta   &    |-  et   &    |-  ze   &    |-  si   &    |-  rh   &    |-  mu   &    |-  la   &    |-  ka   &    |- jph   &    |- jps   &    |- jch   &    |- jth   =>    |-  (
 ( ( ( ( ( ( ( ( ( ( ( ( ( ( ph  /\  ps )  /\  ch )  /\  th )  /\  ta )  /\  et )  /\  ze )  /\  si )  /\  rh )  /\  mu )  /\  la )  /\  ka )  /\ jph )  /\ jps
 )  /\ jch ) 
 -> jth )
 
Theoremdandysum2p2e4 27857

CONTRADICTION PROVED AT 1 + 1 = 2 .

Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses.

Note: Values that when added which exceed a 4bit value are not supported.

Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'.

How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit.

( et <-> F ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit. (Contributed by Jarvin Udandy, 6-Sep-2016.)

 |-  ( ph 
 <->  ( th  /\  ta ) )   &    |-  ( ps  <->  ( et  /\  ze ) )   &    |-  ( ch  <->  ( si  /\  rh ) )   &    |-  ( th  <->  F.  )   &    |-  ( ta  <->  F.  )   &    |-  ( et  <->  T.  )   &    |-  ( ze 
 <->  T.  )   &    |-  ( si  <->  F.  )   &    |-  ( rh  <->  F.  )   &    |-  ( mu  <->  F.  )   &    |-  ( la  <->  F.  )   &    |-  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) )   &    |-  (jph  <->  (
 ( et  \/_  ze )  \/  ph ) )   &    |-  (jps  <->  ( ( si  \/_  rh )  \/  ps ) )   &    |-  (jch  <->  ( ( mu 
 \/_  la )  \/  ch ) )   =>    |-  ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
 ph 
 <->  ( th  /\  ta ) )  /\  ( ps  <->  ( et  /\  ze )
 ) )  /\  ( ch 
 <->  ( si  /\  rh ) ) )  /\  ( th  <->  F.  ) )  /\  ( ta  <->  F.  ) )  /\  ( et  <->  T.  ) )  /\  ( ze  <->  T.  ) )  /\  ( si  <->  F.  ) )  /\  ( rh  <->  F.  ) )  /\  ( mu  <->  F.  ) )  /\  ( la  <->  F.  ) )  /\  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) ) )  /\  (jph  <->  ( ( et  \/_  ze )  \/  ph ) ) ) 
 /\  (jps  <->  (
 ( si  \/_  rh )  \/  ps ) ) ) 
 /\  (jch  <->  (
 ( mu  \/_  la )  \/  ch ) ) ) 
 ->  ( ( ( ( ka  <->  F.  )  /\  (jph  <->  F.  ) )  /\  (jps  <->  T.  ) )  /\  (jch  <->  F.  ) ) )
 
Theoremmdandysum2p2e4 27858 CONTRADICTION PROVED AT 1 + 1 = 2 . Luckily Mario Carneiro did a successful version of his own.

See Mario's Relevant Work: 1.3.14 Half-adders and full adders in propositional calculus

Given the right hypotheses we can prove a dandysum of 2+2=4. The qed step is the value '4' in Decimal BEING IMPLIED by the hypotheses.

Note: Values that when added which exceed a 4bit value are not supported.

Note: Digits begin from left (least) to right (greatest). e.g. 1000 would be '1', 0100 would be '2'. 0010 would be '4'.

How to perceive the hypotheses' bits in order: ( th <-> F. ), ( ta <-> F. ) Would be input value X's first bit, and input value Y's first bit.

( et <-> F. ), ( ze <-> F. ) would be input value X's second bit, and input value Y's second bit.

In mdandysum2p2e4, one might imagine what jth or jta could be then do the math with their truths. Also limited to the restriction jth, jta are having opposite truths equivalent to the stated truth constants.

(Contributed by Jarvin Udandy, 6-Sep-2016.)

 |-  (jth  <->  F.  )   &    |-  (jta  <->  T.  )   &    |-  ( ph  <->  ( th  /\  ta ) )   &    |-  ( ps  <->  ( et  /\  ze ) )   &    |-  ( ch  <->  ( si  /\  rh ) )   &    |-  ( th  <-> jth )   &    |-  ( ta 
 <-> jth
 )   &    |-  ( et  <-> jta )   &    |-  ( ze 
 <-> jta
 )   &    |-  ( si  <-> jth )   &    |-  ( rh 
 <-> jth
 )   &    |-  ( mu  <-> jth )   &    |-  ( la 
 <-> jth
 )   &    |-  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) )   &    |-  (jph  <->  (
 ( et  \/_  ze )  \/  ph ) )   &    |-  (jps  <->  ( ( si  \/_  rh )  \/  ps ) )   &    |-  (jch  <->  ( ( mu 
 \/_  la )  \/  ch ) )   =>    |-  ( ( ( ( ( ( ( ( ( ( ( ( ( ( ( (
 ph 
 <->  ( th  /\  ta ) )  /\  ( ps  <->  ( et  /\  ze )
 ) )  /\  ( ch 
 <->  ( si  /\  rh ) ) )  /\  ( th  <->  F.  ) )  /\  ( ta  <->  F.  ) )  /\  ( et  <->  T.  ) )  /\  ( ze  <->  T.  ) )  /\  ( si  <->  F.  ) )  /\  ( rh  <->  F.  ) )  /\  ( mu  <->  F.  ) )  /\  ( la  <->  F.  ) )  /\  ( ka  <->  ( ( th  \/_ 
 ta )  \/_  ( th  /\  ta ) ) ) )  /\  (jph  <->  ( ( et  \/_  ze )  \/  ph ) ) ) 
 /\  (jps  <->  (
 ( si  \/_  rh )  \/  ps ) ) ) 
 /\  (jch  <->  (
 ( mu  \/_  la )  \/  ch ) ) ) 
 ->  ( ( ( ( ka  <->  F.  )  /\  (jph  <->  F.  ) )  /\  (jps  <->  T.  ) )  /\  (jch  <->  F.  ) ) )
 
19.22  Mathbox for Alexander van der Vekens
 
19.22.1  Double restricted existential uniqueness
 
19.22.1.1  Restricted quantification (extension)
 
Theoremr19.32 27859 Theorem 19.32 of [Margaris] p. 90 with restricted quantifiers, analogous to r19.32v 2846. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  F/ x ph   =>    |-  ( A. x  e.  A  ( ph  \/  ps )  <->  ( ph  \/  A. x  e.  A  ps ) )
 
Theoremrexsb 27860* An equivalent expression for restricted existence, analogous to exsb 2206. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  ph  <->  E. y  e.  A  A. x ( x  =  y  ->  ph ) )
 
Theoremrexrsb 27861* An equivalent expression for restricted existence, analogous to exsb 2206. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  ph  <->  E. y  e.  A  A. x  e.  A  ( x  =  y  ->  ph ) )
 
Theorem2rexsb 27862* An equivalent expression for double restricted existence, analogous to rexsb 27860. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  A. x A. y ( ( x  =  z  /\  y  =  w )  ->  ph )
 )
 
Theorem2rexrsb 27863* An equivalent expression for double restricted existence, analogous to 2exsb 2208. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ( x  =  z  /\  y  =  w )  ->  ph )
 )
 
Theoremcbvral2 27864* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvral2v 2932. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ z ph   &    |-  F/ x ch   &    |-  F/ w ch   &    |-  F/ y ps   &    |-  ( x  =  z  ->  ( ph  <->  ch ) )   &    |-  (
 y  =  w  ->  ( ch  <->  ps ) )   =>    |-  ( A. x  e.  A  A. y  e.  B  ph  <->  A. z  e.  A  A. w  e.  B  ps )
 
Theoremcbvrex2 27865* Change bound variables of double restricted universal quantification, using implicit substitution, analogous to cbvrex2v 2933. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ z ph   &    |-  F/ x ch   &    |-  F/ w ch   &    |-  F/ y ps   &    |-  ( x  =  z  ->  ( ph  <->  ch ) )   &    |-  (
 y  =  w  ->  ( ch  <->  ps ) )   =>    |-  ( E. x  e.  A  E. y  e.  B  ph  <->  E. z  e.  A  E. w  e.  B  ps )
 
Theorem2ralbiim 27866 Split a biconditional and distribute 2 quantifiers, analogous to 2albiim 1622 and ralbiim 2835. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( A. x  e.  A  A. y  e.  B  (
 ph 
 <->  ps )  <->  ( A. x  e.  A  A. y  e.  B  ( ph  ->  ps )  /\  A. x  e.  A  A. y  e.  B  ( ps  ->  ph ) ) )
 
19.22.1.2  The empty set (extension)
 
Theoremraaan2 27867* Rearrange restricted quantifiers with two different restricting classes, analogous to raaan 3727. It is necessary that either both restricting classes are empty or both are not empty. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  F/ y ph   &    |-  F/ x ps   =>    |-  (
 ( A  =  (/)  <->  B  =  (/) )  ->  ( A. x  e.  A  A. y  e.  B  (
 ph  /\  ps )  <->  (
 A. x  e.  A  ph 
 /\  A. y  e.  B  ps ) ) )
 
19.22.1.3  Restricted uniqueness and "at most one" quantification
 
Theoremrmoimi 27868 Restricted "at most one" is preserved through implication (note wff reversal). (Contributed by Alexander van der Vekens, 17-Jun-2017.)
 |-  ( ph  ->  ps )   =>    |-  ( E* x  e.  A ps  ->  E* x  e.  A ph )
 
Theorem2reu5a 27869 Double restricted existential uniqueness in terms of restricted existence and restricted "at most one." (Contributed by Alexander van der Vekens, 17-Jun-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ph  <->  ( E. x  e.  A  ( E. y  e.  B  ph 
 /\  E* y  e.  B ph )  /\  E* x  e.  A ( E. y  e.  B  ph  /\  E* y  e.  B ph ) ) )
 
Theoremreuimrmo 27870 Restricted uniqueness implies restricted "at most one" through implication, analogous to euimmo 2329. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  ( ph  ->  ps )  ->  ( E! x  e.  A  ps  ->  E* x  e.  A ph ) )
 
Theoremrmoanim 27871* Introduction of a conjunct into restricted "at most one" quantifier, analogous to moanim 2336. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  F/ x ph   =>    |-  ( E* x  e.  A ( ph  /\  ps ) 
 <->  ( ph  ->  E* x  e.  A ps ) )
 
Theoremreuan 27872* Introduction of a conjunct into restricted uniqueness quantifier, analogous to euan 2337. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  F/ x ph   =>    |-  ( E! x  e.  A  ( ph  /\  ps ) 
 <->  ( ph  /\  E! x  e.  A  ps ) )
 
19.22.1.4  Analogs to Existential uniqueness (double quantification)
 
Theorem2reurex 27873* Double restricted quantification with existential uniqueness, analogous to 2euex 2352. (Contributed by Alexander van der Vekens, 24-Jun-2017.)
 |-  ( E! x  e.  A  E. y  e.  B  ph 
 ->  E. y  e.  B  E! x  e.  A  ph )
 
Theorem2reurmo 27874* Double restricted quantification with restricted existential uniqueness and restricted "at most one.", analogous to 2eumo 2353. (Contributed by Alexander van der Vekens, 24-Jun-2017.)
 |-  ( E! x  e.  A  E* y  e.  B ph 
 ->  E* x  e.  A E! y  e.  B  ph )
 
Theorem2reu2rex 27875* Double restricted existential uniqueness, analogous to 2eu2ex 2354. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ph 
 ->  E. x  e.  A  E. y  e.  B  ph )
 
Theorem2rmoswap 27876* A condition allowing swap of restricted "at most one" and restricted existential quantifiers, analogous to 2moswap 2355. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  E* y  e.  B ph 
 ->  ( E* x  e.  A E. y  e.  B  ph  ->  E* y  e.  B E. x  e.  A  ph ) )
 
Theorem2rexreu 27877* Double restricted existential uniqueness implies double restricted uniqueness quantification, analogous to 2exeu 2357. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  ->  E! x  e.  A  E! y  e.  B  ph )
 
Theorem2reu1 27878* Double restricted existential uniqueness. This theorem shows a condition under which a "naive" definition matches the correct one, analogous to 2eu1 2360. (Contributed by Alexander van der Vekens, 25-Jun-2017.)
 |-  ( A. x  e.  A  E* y  e.  B ph 
 ->  ( E! x  e.  A  E! y  e.  B  ph  <->  ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph ) ) )
 
Theorem2reu2 27879* Double restricted existential uniqueness, analogous to 2eu2 2361. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  ( E! y  e.  B  E. x  e.  A  ph 
 ->  ( E! x  e.  A  E! y  e.  B  ph  <->  E! x  e.  A  E. y  e.  B  ph ) )
 
Theorem2reu3 27880* Double restricted existential uniqueness, analogous to 2eu3 2362. (Contributed by Alexander van der Vekens, 29-Jun-2017.)
 |-  ( A. x  e.  A  A. y  e.  B  ( E* x  e.  A ph 
 \/  E* y  e.  B ph )  ->  ( ( E! x  e.  A  E! y  e.  B  ph 
 /\  E! y  e.  B  E! x  e.  A  ph )  <->  ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph ) ) )
 
Theorem2reu4a 27881* Definition of double restricted existential uniqueness ("exactly one  x and exactly one  y"), analogous to 2eu4 2363 with the additional requirement that the restricting classes are not empty (which is not necessary as shown in 2reu4 27882). (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  (
 ( A  =/=  (/)  /\  B  =/= 
 (/) )  ->  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  ( E. x  e.  A  E. y  e.  B  ph  /\  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ph  ->  ( x  =  z  /\  y  =  w )
 ) ) ) )
 
Theorem2reu4 27882* Definition of double restricted existential uniqueness ("exactly one  x and exactly one  y"), analogous to 2eu4 2363. (Contributed by Alexander van der Vekens, 1-Jul-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  ( E. x  e.  A  E. y  e.  B  ph  /\  E. z  e.  A  E. w  e.  B  A. x  e.  A  A. y  e.  B  ( ph  ->  ( x  =  z  /\  y  =  w )
 ) ) )
 
Theorem2reu7 27883* Two equivalent expressions for double restricted existential uniqueness, analogous to 2eu7 2366. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  (
 ( E! x  e.  A  E. y  e.  B  ph  /\  E! y  e.  B  E. x  e.  A  ph )  <->  E! x  e.  A  E! y  e.  B  ( E. x  e.  A  ph 
 /\  E. y  e.  B  ph ) )
 
Theorem2reu8 27884* Two equivalent expressions for double restricted existential uniqueness, analogous to 2eu8 2367. Curiously, we can put  E! on either of the internal conjuncts but not both. We can also commute  E! x  e.  A E! y  e.  B using 2reu7 27883. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( E! x  e.  A  E! y  e.  B  ( E. x  e.  A  ph 
 /\  E. y  e.  B  ph )  <->  E! x  e.  A  E! y  e.  B  ( E! x  e.  A  ph 
 /\  E. y  e.  B  ph ) )
 
19.22.2  Alternative definitions of function's and operation's values

The current definition of the value 
( F `  A
) of a function  F for an argument  A (see df-fv 5453) assures that this value is always a set, see fex 5960. This is because this definition can be applied to any classes  F and  A, and evaluates to the empty set when it is not meaningful (as shown by ndmfv 5746 and fvprc 5713).

Although it is very convenient for many theorems on functions and their proofs, there are some cases in which from  ( F `  A
)  =  (/) alone it cannot be decided/derived if  ( F `  A ) is meaningful ( F is actually a function which is defined for  A and really has the function value  (/)) or not. Therefore, additional assumptions are required, such as  (/)  e/  ran  F,  (/)  e.  ran  F or 
Fun  F  /\  A  e. 
dom  F (see, for example, ndmfvrcl 5747).

To avoid such an ambiguity, an alternative definition  ( F''' A ) ( see df-afv 27889) would be possible which evaluates to the universal class ( ( F''' A )  =  _V) if it is not meaningful (see afvnfundmuv 27917, ndmafv 27918, afvprc 27922 and nfunsnafv 27920), and which corresponds to the current definition ( ( F `  A )  =  ( F''' A )) if it is (see afvfundmfveq 27916). That means  ( F''' A )  =  _V  ->  ( F `  A )  =  (/) (see afvpcfv0 27924), but  ( F `  A )  =  (/)  ->  ( F''' A )  =  _V is not generally valid.

In the theory of partial functions, it is a common case that  F is not defined at  A, which also would result in  ( F''' A )  =  _V. In this context we say  ( F''' A ) "is not defined" instead of "is not meaningful".

With this definition the following intuitive equivalence holds:  ( F''' A )  e.  _V <-> " ( F''' A ) is meaningful/defined".

An interesting question would be if 
( F `  A
) could be replaced by  ( F''' A ) in most of the theorems based on function's values. If we look at the (currently 19) proofs using the definition df-fv 5453 of 
( F `  A
), we see that analogons for the following 8 theorems can be proven using the alternative definition: fveq1 5718-> afveq1 27912, fveq2 5719-> afveq2 27913, nffv 5726-> nfafv 27914, csbfv12g 5729-> csbafv12g , fvres 5736-> afvres 27950, rlimdm 12333-> rlimdmafv 27955, tz6.12-1 5738-> tz6.12-1-afv 27952, fveu 5711-> afveu 27931.

3 theorems proved by directly using df-fv 5453 are within a mathbox (fvsb 27569) or not used (isumclim3 12531, avril1 21745).

However, the remaining 8 theorems proved by directly using df-fv 5453 are used more or less often:

* fvex 5733: used in about 1750 proofs.

* tz6.12-1 5738: root theorem of many theorems which have not a strict analogon, and which are used many times: fvprc 5713 (used in about 127 proofs), tz6.12i 5742 (used - indirectly via fvbr0 5743 and fvrn0 5744- in 18 proofs, and in fvclss 5971 used in fvclex 5972 used in fvresex 5973, which is not used!), dcomex 8316 (used in 4 proofs), ndmfv 5746 (used in 86 proofs) and nfunsn 5752 (used by dffv2 5787 which is not used).

* fv2 5714: only used by elfv 5717, which is only used by fv3 5735, which is not used.

* dffv3 5715: used by dffv4 5716 (the previous "df-fv"), which now is only used in deprecated (usage discouraged) theorems or within mathboxes (csbfv12gALT 5730, csbfv12gALTVD 28865), by shftval 11877 (itself used in 9 proofs), by dffv5 25719 (mathbox) and by fvco2 5789, which has the analogon afvco2 27954.

* fvopab5 6525: used only by ajval 22351 (not used) and by adjval 23381 ( used - indirectly - in 9 proofs).

* zsum 12500: used (via isum 12501, sum0 12503 and fsumsers 12510) in more than 90 proofs.

* isumshft 12607: used in pserdv2 20334 and (via logtayl 20539) 4 other proofs.

* ovtpos 6485: used in 14 proofs.

As a result of this analysis we can say that the current definition of a function's value is crucial for Metamath and cannot be exchanged easily with an alternative definition. While fv2 5714, dffv3 5715, fvopab5 6525, zsum 12500, isumshft 12607 and ovtpos 6485 are not critical or are, hopefully, also valid for the alternative definition, fvex 5733 and tz6.12-1 5738 (and the theorems based on them) are essential for the current definition of function values.

With the same arguments, an alternatvie definition of operation's values (( A O B)) could be meaningful to avoid ambiguities, see df-aov 27890.

For additional discussions/material see https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4.

 
Syntaxwdfat 27885 Extend the definition of a wff to include the "defined at" predicate. (Read: (The Function)  F is defined at (the argument)  A). In a previous version, the token "def@" was used. However, since the @ is used (informally) as a replacement for $ in commented out sections that may be deleted some day. While there is no violation of any standard to use the @ in a token, it could make the search for such commented-out sections slightly more difficult. (See remark of Norman Megill at https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4).
 wff  F defAt  A
 
Syntaxcafv 27886 Extend the definition of a class to include the value of a function. (Read: The value of  F at  A, or " F of  A."). In a previous version, the symbol " ' " was used. However, since the similarity with the symbol 
` used for the current definition of a function's value (see df-fv 5453), which, by the way, was intended to visualize that in many cases  ` and " ' " are exchangeable, makes reading the theorems, especially those which uses both definitions as dfafv2 27910, very difficult, 3 apostrophes ''' are used now so that it's easier to distinguish from df-fv 5453 and df-ima 4882. And not three backticks ( three times  ` ) since that would be annoying to escape in a comment. (See remark of Norman Megill and Gerard Lang at https://groups.google.com/forum/#!topic/metamath/cteNUppB6A4).
 class  ( F''' A )
 
Syntaxcaov 27887 Extend class notation to include the value of an operation  F (such as  +) for two arguments  A and  B. Note that the syntax is simply three class symbols in a row surrounded by special parentheses (exclamation mark with underscore) in contrast to the current definition, see df-ov 6075.
 class (( A F B))
 
Definitiondf-dfat 27888 Definition of the predicate that determines if some class  F is defined as function for an argument  A or, in other words, if the function value for some class  F for an argument  A is defined. We say that  F is defined at  A if a  F is a function restricted to the member  A of its domain. (Contributed by Alexander van der Vekens, 25-May-2017.)
 |-  ( F defAt  A  <->  ( A  e.  dom 
 F  /\  Fun  ( F  |`  { A } )
 ) )
 
Definitiondf-afv 27889* Alternative definition of the value of a function,  ( F''' A ), also known as function application. In contrast to  ( F `  A )  =  (/) (see df-fv 5453 and ndmfv 5746),  ( F''' A )  =  _V if F is not defined for A! (Contributed by Alexander van der Vekens, 25-May-2017.)
 |-  ( F''' A )  =  if ( F defAt  A ,  ( iota x A F x ) ,  _V )
 
Definitiondf-aov 27890 Define the value of an operation. In contrast to df-ov 6075, the alternative definition for a function value ( see df-afv 27889) is used. By this, the value of the operation applied to two arguments is the universal class if the operation is not defined for these two arguments. There are still no restrictions of any kind on what those class expressions may be, although only certain kinds of class expressions - a binary operation  F and its arguments  A and  B- will be useful for proving meaningful theorems. (Contributed by Alexander van der Vekens, 26-May-2017.)
 |- (( A F B))  =  ( F'''
 <. A ,  B >. )
 
19.22.2.1  Restricted quantification (extension)
 
Theoremralbinrald 27891* Elemination of a restricted universal quantification under certain conditions. (Contributed by Alexander van der Vekens, 2-Aug-2017.)
 |-  ( ph  ->  X  e.  A )   &    |-  ( x  e.  A  ->  x  =  X )   &    |-  ( x  =  X  ->  ( ps  <->  th ) )   =>    |-  ( ph  ->  (
 A. x  e.  A  ps 
 <-> 
 th ) )
 
19.22.2.2  The universal class (extension)
 
Theoremnvelim 27892 If a class is the universal class it doesn't belong to any class, generalisation of nvel 4334. (Contributed by Alexander van der Vekens, 26-May-2017.)
 |-  ( A  =  _V  ->  -.  A  e.  B )
 
19.22.2.3  Introduce the Axiom of Power Sets (extension)
 
Theoremalneu 27893 If a statement holds for all sets, there is not a unique set for which the statement holds. (Contributed by Alexander van der Vekens, 28-Nov-2017.)
 |-  ( A. x ph  ->  -.  E! x ph )
 
Theoremeu2ndop1stv 27894* If there is a unique second component in an ordered pair contained in a given set, the first component must be a set. (Contributed by Alexander van der Vekens, 29-Nov-2017.)
 |-  ( E! y <. A ,  y >.  e.  V  ->  A  e.  _V )
 
19.22.2.4  Relations (extension)
 
Theoremsbcrel 27895 Distribute proper substitution through a relation predicate. (Contributed by Alexander van der Vekens, 23-Jul-2017.)
 |-  ( A  e.  V  ->  (
 [. A  /  x ].
 Rel  R  <->  Rel  [_ A  /  x ]_ R ) )
 
Theoremcsbdmg 27896 Distribute proper substitution through the domain of a class. (Contributed by Alexander van der Vekens, 23-Jul-2017.)
 |-  ( A  e.  V  ->  [_ A  /  x ]_ dom  B  =  dom  [_ A  /  x ]_ B )
 
Theoremdmmpt2g 27897* Domain of a class given by the "maps to" notation, closed form of dmmpt2 6412. (Contributed by Alexander van der Vekens, 1-Jun-2017.)
 |-  F  =  ( x  e.  A ,  y  e.  B  |->  C )   =>    |-  ( C  e.  V  ->  dom  F  =  ( A  X.  B ) )
 
Theoremeldmressn 27898 Element of the domain of a restriction to a singleton. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( B  e.  dom  ( F  |`  { A } )  ->  B  =  A )
 
Theoremdmressnsn 27899 The domain of a restriction to a singleton is a singleton. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( A  e.  dom  F  ->  dom  ( F  |`  { A } )  =  { A } )
 
Theoremeldmressnsn 27900 The element of the domain of a restriction to a singleton is the element of the singleton. (Contributed by Alexander van der Vekens, 2-Jul-2017.)
 |-  ( A  e.  dom  F  ->  A  e.  dom  ( F  |` 
 { A } )
 )
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